【LeetCode】Rotate Array 旋轉陣列
阿新 • • 發佈:2019-01-04
題目
Rotate an array of n elements to the right by k steps.
For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].
Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.
題目大意
旋轉陣列,給定陣列長度n,旋轉位移k,將陣列元素迴圈右移k位。
思路
三步翻轉法,《程式設計珠璣》一書上有講,步驟如下:
- 將陣列從k位置分為前後兩半;
- 翻轉前半部分;
- 翻轉後半部分;
- 翻轉整個陣列;
圖示如下:
解答
Java程式碼如下,注意遍歷時候的邊界不要出錯:
public void rotate(int[] nums, int k) {
if(k==0 || nums==null ) return;
k = k%nums.length;
int tmp;
for (int i = 0; i < (nums.length-k)/2; i++) {
tmp = nums[i];
nums[i] = nums[nums.length-k-1 -i];
nums[nums.length-k-1-i] = tmp;
}
/*for (int i : nums) {
System.out.print(i+"-");
}
System.out.println();*/
for (int i = 0; i < k/2; i++) {
tmp = nums[i+nums.length-k];
nums[i+nums.length-k] = nums[nums.length-1 -i];
nums[nums.length-1-i] = tmp;
}
/*for (int i : nums) {
System.out.print(i+" ");
}
System.out.println();*/
for (int i = 0; i < nums.length/2; i++) {
tmp = nums[i];
nums[i] = nums[nums.length-1-i];
nums[nums.length-1-i] = tmp;
}
}